Question

Find the area of the region in the first quadrant bounded on the left by the ​y-​axis, below by the line y equals one third x comma above left by yequalsxplus​4, and above right by yequalsminusx squaredplus10.

Answer 1

Answer:

The bounded area  is: $\frac{73}{6}\approx 12.17$

Step-by-step explanation:

Let's start by plotting the functions that enclose the area, so we can find how to practically use integration. Please see attached image where the area in question has been highlighted in light green. The important points that define where the integrations should be performed are also identified with dots in darker green color. These two important points are: (2, 6) and (3, 1)

So we need to perform two separate integrals and add the appropriate areas at the end. The first integral is that of the difference of function y=x+4 minus function y=(1/3)x , and this integral should go from x = 0 to x = 2 (see the bottom left image with the area in red:

$\int\limits^2_0 {x+4-\frac{x}{3} } \, dx =\int\limits^2_0 {\frac{2x}{3} +4} \, dx=\frac{4}{3} +8= \frac{28}{3}$

The next integral is that of the difference between $y=-x^2+10$ and the bottom line defined by: y = (1/3) x. This integration is in between x = 2 and x = 3 (see bottom right image with the area in red:

$\int\limits^3_2 {-x^2+10-\frac{x}{3} } \, dx =-9+30-\frac{3}{2} -(-\frac{8}{3} +20-\frac{2}{3} )=\frac{39}{2} -\frac{50}{3} =\frac{17}{6}$

Now we need to add the two areas found in order to get the total area:

$\frac{28}{3} +\frac{17}{6} =\frac{73}{6}\approx 12.17$